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velikii [3]
4 years ago
8

Una profesora compra 28 manzanas para compartir con sus estudiantes; al día siguiente revisa la cesta con las frutas y ve que se

le han dañado 2/7 del total de manzanas que había comprado. Para reponer las frutas dañadas ella debe comprar
Mathematics
1 answer:
valina [46]4 years ago
6 0

Answer:

La profesora debe comprar 8 manzanas para reponer la cesta de frutas.

Step-by-step explanation:

La profesora debe reponer la cesta de frutas por la cantidad de manzanas que se encuentran dañadas. La cantidad de manzanas dañadas es igual a las dos séptimas partes del total de manzanas. Es decir:

x = \frac{2}{7} \times (28\,manzanas)

x = \frac{56\,manzanas}{7}

x = 8\,manzanas

La profesora debe comprar 8 manzanas para reponer la cesta de frutas.

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HELPPPPPPPPPPP!!!!!!!!! Oh
Luba_88 [7]

Answer:

We conclude that:

3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}

Hence, option B is correct.

Step-by-step explanation:

Given the expression

3\times 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}

solving

3\times 2\begin{pmatrix}-2&3&0\\ \:-4&2&6\\ \:6&-5&6\end{pmatrix}

Scalar Multiplication: Multiply each of the matrix elements by a scalar

=\begin{pmatrix}3\cdot \:2\left(-2\right)&3\cdot \:2\cdot \:3&3\cdot \:2\cdot \:0\\ 3\cdot \:2\left(-4\right)&3\cdot \:2\cdot \:2&3\cdot \:2\cdot \:6\\ 3\cdot \:2\cdot \:6&3\cdot \:2\left(-5\right)&3\cdot \:2\cdot \:6\end{pmatrix}

Simplify each element

=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}

Therefore, we conclude that:

3\cdot \:2\begin{pmatrix}-2&3&0\\ -4&2&6\\ 6&-5&6\end{pmatrix}=\begin{pmatrix}-12&18&0\\ -24&12&36\\ 36&-30&36\end{pmatrix}

Hence, option B is correct.

5 0
3 years ago
Tres fracciones que representen el 3
FrozenT [24]
Creo que las respuestas son 6/2, 18/6 y 12/4
7 0
3 years ago
Read 2 more answers
I really don't get this im so confused
Tanzania [10]

Answer:

Step-by-step explanation:

m<ABC=90 because of Angle inscribed in Semicircle Theorem

6 0
4 years ago
Read 2 more answers
Suppose the FAA weighed a random sample of 20 airline passengers during the summer and found their weights to have a sample mean
Orlov [11]

Answer:

Step-by-step explanation:

Given Parameters

Mean, x = 180

total samples, n = 20

Standard dev, \sigma = 30

\alpha = 1 - 0.95 = 0.05 at 95% confidence level

Df = n - 1 = 20 - 1 = 19

Critical Value, t_\alpha, is given by

t_{c}=t_{\alpha, df} = t_{0.05,19} = 1.729

a).

Confidence Interval, \mu, is given by the formula

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 1.729 \times \frac{30}{\sqrt{20} }

\mu = 180 +/-11.5985

191.5985 > \mu > 168.4015

b).

Critical Value, t_{\alpha/2}, is given by

t_{c}=t_{\alpha/2, df} = t_{0.05/2,19} = 2.093

Confidence Interval, \mu, is given by

\mu = x +/- t_c \times \frac{s}{\sqrt{n} }

\mu = 180 +/- 2.093 \times \frac{30}{\sqrt{20} }

    = 180 +/- 14.0403

    = 165.9597 < \mu < 194.0403

8 0
3 years ago
G+3&gt;6<br> A (3)<br> B (4)<br> C (-3)<br> D (-4)
aleksley [76]
G>6-3
G>3 is definitely the answer
7 0
3 years ago
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